MHB Justine's question at Yahoo Answers regarding the Midpoint and Trapezoidal Rules

[MarkFL]

Here is the question:

Trapezoidal and Midpoint Estimates?


Show that (1/2)(Tn+Mn)=T2n

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Justine,

First, let's look at the definitions of the two rules:

Midpoint Rule:

The Midpoint Rule is the approximation $$\int_a^b f(x)\,dx\approx M_n$$, where

$$M_n=\frac{b-a}{n}\left[f\left(\frac{x_0+x_1}{2} \right)+f\left(\frac{x_1+x_2}{2} \right)+\cdots+f\left(\frac{x_{n-1}+x_{n}}{2} \right) \right]$$

Trapezoidal Rule:

The Trapezoidal Rule is the approximation $$\int_a^b f(x)\,dx\approx T_n$$, where

$$T_n=\frac{b-a}{2n}\left[f\left(x_0 \right)+2f\left(x_1 \right)+\cdots+2f\left(x_{n-1} \right)+f\left(x_{n} \right) \right]$$

We can divide the interval $[a,b]$ into $2n$ equally spaced partitions, as so we may then write:

$$M_n=\frac{b-a}{n}\left[f\left(\frac{x_0+x_2}{2} \right)+f\left(\frac{x_2+x_4}{2} \right)+\cdots+f\left(\frac{x_{2n-2}+x_{2n}}{2} \right) \right]$$

And so using the midpoint formula, this becomes:

$$M_n=\frac{b-a}{2n}\left[2f\left(x_1 \right)+2f\left(x_3 \right)+\cdots+2f\left(x_{2n-1} \right) \right]$$

Likewise, the Trapezoidal Rule may now be written:

$$T_n=\frac{b-a}{2n}\left[f\left(x_0 \right)+2f\left(x_2 \right)+\cdots+2f\left(x_{2n-2} \right)+f\left(x_{2n} \right) \right]$$

Multiplying both by $$\frac{1}{2}$$ and adding, we find:

$$\frac{1}{2}\left(T_n+M_n \right)=\frac{b-a}{2(2n)}\left[f\left(x_0 \right)+2f\left(x_1 \right)+\cdots+2f\left(x_{2n-1} \right)+f\left(x_{2n} \right) \right]$$

And using the definition of the Trapezoidal Rule, we find:

$$\frac{1}{2}\left(T_n+M_n \right)=T_{2n}$$